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bija089 [108]
3 years ago
15

Write the equation of the line that passes through the given points. (0.-6) and (7.0)

Mathematics
1 answer:
ahrayia [7]3 years ago
6 0
<h2>Answer:    y =  ⁶/₇ ( x  -  7 )</h2>

Step-by-step explanation:

The slope of tghe line (m) = (y₂ - y₁) ÷ (x₂ - x₁)

                                          =  (- 6 -0) ÷ ( 0 - 7)

                                          =    ⁶/₇

We can now use the point-slope form to write the equation for this line:

y - y₁ = m(x - x₁) where (x₁ , y₁) = (7 , 0)

              y =  ⁶/₇ ( x  -  7 )

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Statement ∠T and ∠S are complementary ∠S and ∠U are complementary m∠T+m∠S=90 m∠S+m∠U=90 m∠T+m∠S=m∠S+m∠U m∠U=m∠T ∠U≅∠T
Leto [7]

Answer:

See Explanation

Step-by-step explanation:

According to the Question,

  • Given That,  ∠T and ∠S are complementary & ∠S and ∠U are complementary

Thus, m∠T + m∠S = 90 ------ Equation 1

&  m∠S + m∠U = 90  ------- Equation 2

  • Now, Subtract Equation 2 From Equation 1, we get

m∠T - m∠U = 0

Thus, m∠T = m∠U

The measure of angle T is equal to angle U.(∠U≅∠T).

7 0
3 years ago
You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along
sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}

Then, the time (speed divided by distance) is:

t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1

To optimize this function we have to derive and equal to zero:

\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\  \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\

x

As d_b=x, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

7 0
4 years ago
Plz help me i'll give points and brainliest
RoseWind [281]

Step-by-step explanation:

-1,6 is -1 left 6 up

1,-6 is 1 right -6 down

6,-1 is 6 right -1 down

3 0
3 years ago
Find the vertices and foci of the hyperbola with equation quantity x plus 1 squared divided by 225 minus the quantity of y plus
miss Akunina [59]

Answer:

The vertices are (-16,-5) and (14,-5)

The foci are (-26,-5) and (24,-5)

6 0
4 years ago
PLEASE HELP ME I AM HORRIBLE AT TRIGONOMETRY!!! TRIANGLES ARE SO HARD!!!
OleMash [197]

Answer:

A construction worker may use trigonometry when building the roof of a house. They may have to calculate the right angle that hypotenuse must sit at so water doesn't build up on the roof and cause leaks.

(this may be more of an architect's job, as they are the ones that design the houses)

or

An air traffic controller may use trigonometry to find how long it will take an airplane to land/how far the airplane is from the ground while it is in the sky (hypotenuse), based on its height from the ground (opposite) and the distance away from the runway it is (adjacent). They can also calculate the angle of elevation of the airplane from the tower they work in.

8 0
3 years ago
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